Engineering
Mathematics
Conditional Probability
Question

If P(E1) = 0.2, P(E2) = 0.4 and P(E3) = 0.6 and E1, E2 and E3 are independent events, then the probability that at least one of these events E1, E2 and E3 occurs is

1.2

0.048

0.808

0.952

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Solution
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P(at least one event E1, E2 and E3 occurs)

= P(E1∪E2∪E3) = P(E1) + P(E2) + P(E3) − P(E1∩E2) − P(E2∩E3) − P(E3∩E1) + P(E1∩E2∩E3)

= P(E1) + P(E2) + P(E3) − P(E1)P(E2) − P(E2)P(E3) − P(E3)P(E1) + P(E1)P(E2)P(E3)

= 0.2 + 0.4 + 0.6 − (0.2)(0.4) − (0.4)(0.6) − (0.6)(0.2) + (0.2)(0.4)(0.6)

= 0.808

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